Assume that the continuum is $\aleph_2$ in our ground model $V$. Suppose that $(P_n \, : \, n \in \omega)$ is a fully supported iteration of length $\omega$. Suppose further that the factors of the iteration are proper and add for every step $\omega_1$-many Cohen reals over the previous model.

If we let $P_{\omega}$ be the inverse limit, then there will be new reals which were not added by one of the $P_n$'s. Will there be a real $r$ in $V^{P_{\omega}}$, such that $r$ is Cohen over every $V^{P_n}$?

I am also highly interested in the dual situation, where we replace Cohen forcing with Random forcing in the above. Can we guarantee that there is a real $r$ which is Random over all $V^{P_n}$'s? If not, is it possible to feed in countably many proper factors to the iteration, such that we can exclude reals which are Random/Cohen over the $V^{P_n}$'s

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    $\begingroup$ Do you mean specifically that the forcing at stage $n$ is the forcing $\text{Add}(\omega,\omega_1)$? (After all, even adding one Cohen real will add continuum many Cohen reals to the ground model.) $\endgroup$ Jul 4, 2018 at 23:38
  • $\begingroup$ Yes, I recall learning that from one your answers here. I picked $Add(\omega, \omega_1)$ as it is closer to the actual problem I am facing. Plus it rules out some tricks involving rearranging of the factors. $\endgroup$ Jul 5, 2018 at 5:27
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    $\begingroup$ It is a bit unusual to start with $\mathfrak c=\aleph_2$. There is a perfect antichain in $P_\omega$, so the forcing $P_\omega$ will not have the $\aleph_2$-cc. (And iterating it for $\omega_1$ many steps will collapse $\aleph_2$. But apparently you do not want to continue the iteration beyond $\omega$., so perhaps this is not a problem.) $\endgroup$
    – Goldstern
    Jul 8, 2018 at 15:15
  • $\begingroup$ I should probably add that the motivation for this question has nothing to do with questions concerning cardinal characteristics, so I am not worried about $\aleph_2$ being collapsed. An idea how to solve the problem under CH would be helpful as well. $\endgroup$ Jul 8, 2018 at 17:01
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    $\begingroup$ @Ashutosh It's possible that $r$ is contained in some meager set coded in the ground-model, in which case $r$ is not Cohen-generic over $V$ (and consequently isn't Cohen-generic over any of the $V^{P_n}$.) $\endgroup$
    – Not Mike
    Jul 8, 2018 at 18:52


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